Author Topic: HW1 Problem 6 (41)  (Read 577 times)

Kexin Wang

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HW1 Problem 6 (41)
« on: January 27, 2022, 09:12:08 PM »
\begin{align}
&\left\{\begin{aligned}
&u_{xy}=0,\\[2pt]
&u_{xz}=0;
\end{aligned}\right.\\[2pt]
\end{align}

For this problem I'm not sure if my approach was correct.
\begin{align}
u_{x}=f(x,z)  \textrm{   from the first equation}
\end{align}
\begin{align}
&u_{x}=g(x,y)\textrm{   from the second equation}
\end{align}
\begin{align}
\textrm{Therefore we must have   }\textrm{   }u_{x} = h(x)
\end{align}
\begin{align}
&u=H(x) + m(y,z)\textrm{ } \textrm{where}\textrm{ } H'(x)=h(x)
\end{align}

Victor Ivrii

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Re: HW1 Problem 6 (41)
« Reply #1 on: January 28, 2022, 01:41:36 AM »
Yes, it is correct: $u=H(x)+m(y,z)$ where $H$ and $m$ are arbitrary functions.